实验室通宵刷题模式

2013-7-26  19：00

一百道！！！！

pre：python 写的EOJ比较题的工具

import urllib2import sgmllibimport sysimport stringclass LinksParser(sgmllib.SGMLParser):urls = []def do_a(self, attrs):for name, value in attrs:if value.startswith('problem.php?problemid='):s = value[22:]t = string.atoi(s)self.urls.append(t)#print s,t'''if name == 'div' and value not in self.urls:if value.startswith('http'):self.urls.append(value)#print valueelse:continuereturn'''#return urlsif __name__ == "__main__":# str = ""# if str.strip() is '':# print "str is None"# else:# print "str is no None"myid = 'rgtjf'otherid =  '10072130237'url = 'http://acm.cs.ecnu.edu.cn/userinfo.php?username='problemurl = 'http://acm.cs.ecnu.edu.cn/problem.php?problemid='mysets = []othersets = []notsolved = []filename = 'file'                #输入id'''myid = raw_input('myid')otherid = raw_input('otherid')'''#shuchu chong dingxiangtemp0 = sys.stdoutsys.stdout = open(filename,'w')temp1 = sys.stdout        #temp0为console        #temp1为file        sys.stdout = temp0p =  LinksParser()try:f = urllib2.urlopen(url+myid)except URLError,e:print e.reason()value = f.read()p.feed(value)mysets = p.urls[:]try:f = urllib2.urlopen(url+otherid)except URLError,e:print e.reason()value = f.read()p.feed(value)othersets = p.urls[:]#print mysets#print othersetsc = 0for i in othersets:#print iif i not in mysets:print problemurl+str(i)c = c + 1print 'All:',print cf.close()p.close()

version 0.1

直接打开第一个不同的链接

import urllib2import sgmllibimport sysimport stringimport webbrowserclass LinksParser(sgmllib.SGMLParser):urls = []def do_a(self, attrs):for name, value in attrs:if value.startswith('problem.php?problemid='):s = value[22:]t = string.atoi(s)self.urls.append(t)#print s,t'''if name == 'div' and value not in self.urls:if value.startswith('http'):self.urls.append(value)#print valueelse:continuereturn'''#return urlsif __name__ == "__main__":# str = ""# if str.strip() is '':# print "str is None"# else:# print "str is no None"myid = 'rgtjf'otherid =  '10122130152'url = 'http://acm.cs.ecnu.edu.cn/userinfo.php?username='problemurl = 'http://acm.cs.ecnu.edu.cn/problem.php?problemid='mysets = []othersets = []notsolved = []filename = 'file''''myid = raw_input('myid')otherid = raw_input('otherid')'''#shuchu chong dingxiangtemp0 = sys.stdoutsys.stdout = open(filename,'w')temp1 = sys.stdoutsys.stdout = temp0p =  LinksParser()f = urllib2.urlopen(url+myid)value = f.read()p.feed(value)mysets = p.urls[:]f = urllib2.urlopen(url+otherid)value = f.read()p.feed(value)othersets = p.urls[:]#print mysets#print othersetssaveid = 0;c = 0for i in othersets:#print iif i not in mysets:if saveid == 0:saveid = iprint problemurl+str(i)c = c + 1print 'All:',print cif saveid != 0:webbrowser.open_new_tab(problemurl+str(saveid))'''if saveid != 0:f = urllib2.urlopen(problemurl+str(saveid))value = f.read()print value'''f.close()p.close()

1：http://acm.cs.ecnu.edu.cn/problem.php?problemid=1011

#include <cstdio>#include <deque>#include <queue>#include <set>#include <string>#include <map>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <cmath> using namespace std;typedef long long LL;const int maxn = 100005;int visited[maxn];int z,i,m,l;int main(){int cas = 1;while(scanf("%d%d%d%d",&z,&i,&m,&l),z||i||m||l){memset(visited,0,sizeof(visited));int c = 1;visited[l] = c++;printf("Case %d: ",cas++);while(1){int nl = (z*l+i)%m;//printf("%d\n",nl);if(visited[nl]!=0){printf("%d\n",c-visited[nl]);break;}else {visited[nl] = c++;}l = nl;}}}

2:http://acm.cs.ecnu.edu.cn/problem.php?problemid=1014

#include <cstdio>#include <deque>#include <queue>#include <set>#include <string>#include <map>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <cmath> using namespace std;typedef long long LL;const int maxn = 10;int x[maxn];int y[maxn];int a[maxn];int b[maxn];bool solve(){bool visited[10];memset(visited,false,sizeof(visited));int top = 0;for(int i=0;i<6;i++){if(visited[i] == true)continue;for(int j=i+1;j<6;j++){if(visited[j] == true)continue;if(x[i] == x[j] && y[i] == y[j]){visited[j] = true;a[top] = x[i];b[top] = y[i];top ++;break;}}}if(top != 3)return false;int ta,tb;if(a[0] == a[1]){if(b[0] == a[2]){if(b[1] == b[2])return true;}else if(b[0] == b[2]){if(b[1] == a[2])return true;}}else if(a[0] == b[1]){if(b[0] == a[2]){if(a[1] == b[2])return true;}else if(b[0] == b[2]){if(a[1] == a[2])return true;}}if(a[0] == a[2]){if(b[0] == a[1]){if(b[1] == b[2])return true;}else if(b[0] == b[1]){if(b[2] == a[1])return true;}}else if(a[0] == b[2]){if(b[0] == a[1]){if(a[2] == b[1])return true;}else if(b[0] == b[1]){if(a[1] == a[2])return true;}}return false;}int main(){int t1,t2;while(scanf("%d%d",&t1,&t2)!=-1){if(t1 < t2){x[0] = t1;y[0] = t2;}else{x[0] = t2;y[0] = t1;}for(int i=1;i<6;i++){scanf("%d%d",&t1,&t2);if(t1 < t2){x[i] = t1;y[i] = t2;}else{x[i] = t2;y[i] = t1;}}if(solve())printf("POSSIBLE\n");else printf("IMPOSSIBLE\n");}return 0;}

3.http://acm.cs.ecnu.edu.cn/problem.php?problemid=1018

#include <cstdio>#include <deque>#include <queue>#include <set>#include <string>#include <map>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <cmath> using namespace std;typedef long long LL;const int maxn = 257;int dp[maxn];int main(){int T;scanf("%d",&T);while(T --){string str;cin >> str;int m;cin >> m;vector<int> v[maxn];memset(dp,0x3f,sizeof(dp));while(m --){string patten;cin >> patten;size_t found = str.find(patten);int length = patten.length();while(found!= string::npos){    //cout << found << '\n';    v[found+length-1].push_back(found-1);    //printf("%d -> %d\n",found+length-1,found);    found = str.find(patten,found+1);    //int f = (int)found;}}int length = str.length();for(int i=0;i<length;i++){for(int j=0;j<v[i].size();j++){int t = v[i][j];//printf("%d -> %d\n",i,t);if(t == -1){dp[i] = 1;continue;}if(dp[t] + 1 < dp[i]){dp[i] = dp[t] + 1;}}//printf("%d %d\n",i, dp[i]);}printf("%d\n",dp[length-1]);}}

4：http://acm.cs.ecnu.edu.cn/problem.php?problemid=1034

#include <cstdio>#include <deque>#include <queue>#include <set>#include <string>#include <map>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <cmath> using namespace std;typedef long long LL;const int maxn = 1000005;const double eps = 1e-8;double f[maxn];int d[maxn];void init(){f[1] = 0;for(int i=1;i<maxn;i++){f[i] = f[i-1] + log2(1.0*i);d[i] = ceil(f[i]+eps);//printf("%.2lf\n",f[i]);}}int bs(double x){int l = 1;int r = maxn;int ans;while(l <= r){int m = (l+r) >> 1;if(f[m] < x){ans = m;l = m + 1;}else{r = m - 1;}}return ans;}int pow2(int n){int ans = 1;for(int i=0;i<n;i++){ans *= 2;}return ans;}int main(){//int t = ceil(10+eps);//printf("%d\n",t);init();int n;while(cin >> n, n){n = (n - 1960) / 10;n = pow2(n+2);cout << bs(1.0*n) << endl;}return 0;}

5:http://acm.cs.ecnu.edu.cn/problem.php?problemid=1038

#include <cstdio>#include <deque>#include <queue>#include <set>#include <string>#include <map>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <cmath> using namespace std;typedef long long LL;const double eps = 1e-8;const int maxn = 105;int main(){int n,m;while(scanf("%d\n",&n),n){scanf("%d",&m);set<int> s;for(int i=0;i<n;i++){int t;scanf("%d",&t);s.insert(t);}bool ans = true;while(m --){int c;int r;scanf("%d%d",&c,&r);int cr = 0;for(int i=0;i<c;i++){int t;scanf("%d",&t);if(s.find(t) != s.end()){cr ++;}}if(cr >= r);else ans = false;}if(ans)printf("yes\n");else printf("no\n");}return 0;}

 

6：http://acm.cs.ecnu.edu.cn/problem.php?problemid=1035

#include <cstdio>#include <deque>#include <queue>#include <set>#include <string>#include <map>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <cmath> using namespace std;typedef long long LL;const double eps = 1e-8;const int maxn = 1005;bool v[maxn];bool visited[maxn];int n,z,m;bool solve(int k){memset(visited,false,sizeof(visited));int t = 0;//visited[t] = true;while(1){if(t == z)return true;if(visited[t] == true)return false;if(v[t] == true)return false;visited[t] = true;t = (t+k)%n;}}int main(){while(scanf("%d%d%d",&n,&z,&m)!=-1){memset(v,false,sizeof(v));z --;while(m --){int t;scanf("%d",&t);t --;v[t] = true;}int k=1;while(1){if(solve(k)){printf("%d\n",k);break;}k++;}}return 0;}

7.http://acm.cs.ecnu.edu.cn/problem.php?problemid=1044

#include <cstdio>#include <deque>#include <set>#include <string>#include <map>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>using namespace std;typedef long long LL;const int maxn = 100005;struct node{int v;int next;int val;}e[maxn*100];int head[maxn];int tot;bool vis[maxn];int d[maxn];void init(){tot = 0;memset(head,-1,sizeof(head));}void add(int u,int v,int val){e[tot].v = v;e[tot].val = val;e[tot].next = head[u];head[u] = tot ++;}void spfa(int st){memset(vis,false,sizeof(vis));memset(d,0x3f,sizeof(d));queue<int> q;q.push(st);vis[st] = true;d[st] = 0;while(!q.empty()){int u = q.front();q.pop();vis[u] = false;for(int i=head[u];i!=-1;i=e[i].next){int v = e[i].v;if(d[u]+e[i].val < d[v]){d[v] = d[u] + e[i].val;if(!vis[v]){vis[v] = true;q.push(v);}}}}}int main(){int n;while(~scanf("%d",&n)){init();for(int i=0;i<n;i++){int s,e,l,c;l = 1;scanf("%d%d",&s,&c);while(c --){scanf("%d",&e);add(s,e,l);add(e,s,l);}}int st,ed;scanf("%d%d",&st,&ed);spfa(st);printf("%d %d %d\n",st,ed,d[ed]-1);}return 0;}

EOJ 内存还能开这么大啊，不容易！

8：http://acm.cs.ecnu.edu.cn/problem.php?problemid=1047

#include <cstdio>#include <deque>#include <set>#include <string>#include <map>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>using namespace std;typedef long long LL;/* **************************************************************************//二分图匹配（匈牙利算法的DFS实现）//初始化：g[][]两边顶点的划分情况//建立g[i][j]表示i->j的有向边就可以了，是左边向右边的匹配//g没有边相连则初始化为0//uN是匹配左边的顶点数，vN是匹配右边的顶点数//调用：res=hungary();输出最大匹配数//优点：适用于稠密图，DFS找增广路，实现简洁易于理解//时间复杂度:O(VE)//***************************************************************************///顶点编号从0开始的const int MAXN=510;int uN,vN;//u,v数目int g[MAXN][MAXN];int linker[MAXN];bool used[MAXN];bool dfs(int u)//从左边开始找增广路径{    int v;    for(v=0;v<vN;v++)//这个顶点编号从0开始，若要从1开始需要修改if(g[u][v]&&!used[v]){    used[v]=true;    if(linker[v]==-1 || dfs(linker[v]))    {//找增广路，反向        linker[v]=u;        return true;    }}    return false;//这个不要忘了，经常忘记这句}int hungary(){    int res=0;    int u;    memset(linker,-1,sizeof(linker));    for(u=0;u<uN;u++)    {        memset(used,0,sizeof(used));        if(dfs(u)) res++;    }    return res;}struct node{int h;char sex[4];char style[105];char sport[105];}e[MAXN];int main(){int T;scanf("%d",&T);while(T --){int n;scanf("%d",&n);uN = vN = n;for(int i=0;i<n;i++){cin >> e[i].h >> e[i].sex >> e[i].style >> e[i].sport;}memset(g,0,sizeof(g));for(int i=0;i<n;i++){for(int j=i+1;j<n;j++){if(i == j)continue;if(abs(e[i].h-e[j].h)>40          \|| strcmp(e[i].sex,e[j].sex) == 0 \|| strcmp(e[i].style,e[j].style) != 0  \|| strcmp(e[i].sport,e[j].sport) == 0){;}else{g[i][j] = 1;g[j][i] = 1;}  }}int ans = hungary();printf("%d\n",uN-ans/2);}return 0;}

果然要建双向边啊，单向边果然有问题！

9：http://acm.cs.ecnu.edu.cn/problem.php?problemid=1050

#include<stdio.h>#define L 110#define LL 10int N,n,i,j,k,a[L][L],b[L][L],c[L][L];int a11[LL][L][L],a12[LL][L][L],a21[LL][L][L],a22[LL][L][L],b11[LL][L][L],b12[LL][L][L],b21[LL][L][L],b22[LL][L][L],c11[LL][L][L],c12[LL][L][L],c21[LL][L][L],c22[LL][L][L];int d1[LL][L][L],d2[LL][L][L],d3[LL][L][L],d4[LL][L][L],d5[LL][L][L],d6[LL][L][L],d7[LL][L][L],t1[LL][L][L],t2[LL][L][L];void fen(int a[L][L],int a11[L][L],int a12[L][L],int a21[L][L],int a22[L][L],int n) {    for(i=1;i<=n;++i) {for(j=1;j<=n;++j) {a11[i][j]=a[i][j];            a12[i][j]=a[i][j+n];            a21[i][j]=a[i+n][j];            a22[i][j]=a[i+n][j+n];        }    }} void he(int a[L][L],int a11[L][L],int a12[L][L],int a21[L][L],int a22[L][L],int n) {    int i,j;    for (i=1;i<=n;++i) {        for (j=1;j<=n;++j) {a[i][j]=a11[i][j];            a[i][j+n]=a12[i][j];            a[i+n][j]=a21[i][j];            a[i+n][j+n]=a22[i][j];        }    }} void add(int c[L][L],int a[L][L],int b[L][L],int n) {    int i, j;    for (i=1;i<=n;++i)         for (j=1;j<=n;++j) c[i][j]=a[i][j]+b[i][j];} void sub(int c[L][L],int a[L][L],int b[L][L],int n) {    int i,j;    for(i=1;i<=n;++i)         for (j=1;j<=n;++j)c[i][j]=a[i][j]-b[i][j];}void mul(int c[L][L],int a[L][L],int b[L][L],int n, int dep) {           if (n==1)     {        c[1][1]=a[1][1]*b[1][1];        }    else if(n>1&&n<=100)    {fen(a,a11[dep],a12[dep],a21[dep],a22[dep],n/2);fen(b,b11[dep],b12[dep],b21[dep],b22[dep],n/2);sub(t1[dep],b12[dep],b22[dep],n/2);mul(d1[dep],a11[dep],t1[dep],n/2,dep+1);add(t2[dep],a11[dep],a12[dep],n/2);mul(d2[dep],t2[dep],b22[dep],n/2,dep+1); add(t1[dep],a21[dep],a22[dep],n/2);mul(d3[dep],t1[dep],b11[dep],n/2,dep+1); sub(t2[dep],b21[dep],b11[dep],n/2);mul(d4[dep],a22[dep],t2[dep],n/2,dep+1);add(t1[dep],a11[dep],a22[dep],n/2);add(t2[dep],b11[dep],b22[dep],n/2);mul(d5[dep],t1[dep],t2[dep],n/2,dep+1); sub(t1[dep],a12[dep],a22[dep],n/2);add(t2[dep],b21[dep],b22[dep],n/2);mul(d6[dep],t1[dep],t2[dep],n/2,dep+1);sub(t1[dep],a11[dep],a21[dep],n/2);add(t2[dep],b11[dep],b12[dep],n/2);mul(d7[dep],t1[dep],t2[dep],n/2,dep+1);     add(t1[dep],d4[dep],d5[dep],n/2);sub(t2[dep],d2[dep],d6[dep],n/2); sub(c11[dep],t1[dep],t2[dep],n/2); add(c12[dep],d1[dep],d2[dep],n/2); add(c21[dep],d3[dep],d4[dep],n/2); add(t1[dep],d1[dep],d5[dep],n/2);add(t2[dep],d3[dep],d7[dep],n/2);sub(c22[dep],t1[dep],t2[dep],n/2);he(c,c11[dep],c12[dep],c21[dep],c22[dep],n/2);}} void main() {        scanf("%d",&N);        for(k=1;k<=N;k++)         {                            scanf("%d",&n);            for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&a[i][j]);for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&b[i][j]);            mul(c,a,b,n,1);            for(i=1;i<=n;i++){for(j=1;j<n;j++)printf("%d ",c[i][j]);printf("%d",c[i][j]);printf("\n");}        }        }

 

分治算法，具体见：http://www.cnblogs.com/rollenholt/archive/2011/08/27/2155854.html

10.http://acm.cs.ecnu.edu.cn/problem.php?problemid=1062

//============================================================================// Name        : 基本函数模板.cpp// Author      : // Version     :// Copyright   : Your copyright notice// Description : Hello World in C++, Ansi-style//============================================================================#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#include <map>#include <vector>#include <set>#include <string>#include <math.h>using namespace std;const double eps = 1e-8;const double PI = acos(-1.0);int sgn(double x){if(fabs(x) < eps)return 0;if(x < 0)return -1;else return 1;}struct Point{double x,y;Point(){}Point(double _x,double _y){x = _x;y = _y;}Point operator -(const Point &b)const{return Point(x - b.x,y - b.y);}//叉积double operator ^(const Point &b)const{return x*b.y - y*b.x;}//点积double operator *(const Point &b)const{return x*b.x + y*b.y;}//绕原点旋转角度B（弧度值），后x,y的变化void transXY(double B){double tx = x,ty = y;x = tx*cos(B) - ty*sin(B);y = tx*sin(B) + ty*cos(B);}};struct Line{Point s,e;Line(){}Line(Point _s,Point _e){s = _s;e = _e;}//两直线相交求交点//第一个值为0表示直线重合，为1表示平行，为0表示相交,为2是相交//只有第一个值为2时，交点才有意义    pair<int,Point> operator &(const Line &b)const    {        Point res = s;        if(sgn((s-e)^(b.s-b.e)) == 0)        {            if(sgn((s-b.e)^(b.s-b.e)) == 0)                return make_pair(0,res);//重合            else return make_pair(1,res);//平行        }        double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));        res.x += (e.x-s.x)*t;        res.y += (e.y-s.y)*t;        return make_pair(2,res);    }};//*两点间距离double dist(Point a,Point b){    return sqrt((a-b)*(a-b));}//*判断线段相交bool inter(Line l1,Line l2){return    max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&    max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&    max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&    max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&    sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0 &&    sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= 0;}//判断直线和线段相交bool Seg_inter_line(Line l1,Line l2) //判断直线l1和线段l2是否相交{return sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0;}//点到直线距离//返回为result,是点到直线最近的点Point PointToLine(Point P,Line L){    Point result;    double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));    result.x = L.s.x + (L.e.x-L.s.x)*t;    result.y = L.s.y + (L.e.y-L.s.y)*t;    return result;}//点到线段的距离//返回点到线段最近的点Point NearestPointToLineSeg(Point P,Line L){    Point result;    double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));    if(t >= 0 && t <= 1)    {        result.x = L.s.x + (L.e.x - L.s.x)*t;        result.y = L.s.y + (L.e.y - L.s.y)*t;    }    else    {        if(dist(P,L.s) < dist(P,L.e))            result = L.s;        else result = L.e;    }    return result;}//计算多边形面积//点的编号从0~n-1double CalcArea(Point p[],int n){    double res = 0;    for(int i = 0;i < n;i++)        res += (p[i]^p[(i+1)%n])/2;    return fabs(res);}//*判断点在线段上bool OnSeg(Point P,Line L){    return    sgn((L.s-P)^(L.e-P)) == 0 &&    sgn((P.x - L.s.x) * (P.x - L.e.x)) <= 0 &&    sgn((P.y - L.s.y) * (P.y - L.e.y)) <= 0;}//*判断点在凸多边形内//点形成一个凸包，而且按逆时针排序（如果是顺时针把里面的<0改为>0）//点的编号:0~n-1//返回值：//-1:点在凸多边形外//0:点在凸多边形边界上//1:点在凸多边形内int inConvexPoly(Point a,Point p[],int n){for(int i = 0;i < n;i++){if(sgn((p[i]-a)^(p[(i+1)%n]-a)) < 0)return -1;else if(OnSeg(a,Line(p[i],p[(i+1)%n])))return 0;}return 1;}//*判断点在任意多边形内//射线法，poly[]的顶点数要大于等于3,点的编号0~n-1//返回值//-1:点在凸多边形外//0:点在凸多边形边界上//1:点在凸多边形内int inPoly(Point p,Point poly[],int n){int cnt;Line ray,side;cnt = 0;ray.s = p;ray.e.y = p.y;ray.e.x = -100000000000.0;//-INF,注意取值防止越界for(int i = 0;i < n;i++){side.s = poly[i];side.e = poly[(i+1)%n];if(OnSeg(p,side))return 0;//如果平行轴则不考虑if(sgn(side.s.y - side.e.y) == 0)continue;if(OnSeg(side.s,ray)){if(sgn(side.s.y - side.e.y) > 0)cnt++;}else if(OnSeg(side.e,ray)){if(sgn(side.e.y - side.s.y) > 0)cnt++;}else if(inter(ray,side))cnt++;}if(cnt % 2 == 1)return 1;else return -1;}//判断凸多边形//允许共线边//点可以是顺时针给出也可以是逆时针给出//点的编号0~n-1bool isconvex(Point poly[],int n){bool s[3];memset(s,false,sizeof(s));for(int i = 0;i < n;i++){s[sgn( (poly[(i+1)%n]-poly[i])^(poly[(i+2)%n]-poly[i]) )+1] = true;if(s[0] && s[2])return false;}return true;}int main(){Line l1,l2;while(cin >> l1.s.x >> l1.s.y \      >> l1.e.x >> l1.e.y \      >> l2.s.x >> l2.s.y \      >> l2.e.x >> l2.e.y){if(inter(l1,l2))printf("Yes\n");else printf("No\n");}return 0;}

模板，不解释！

11.http://acm.cs.ecnu.edu.cn/problem.php?problemid=1051

    #include<iostream>    #include<vector>    using namespace std;    typedef long long LL;    inline LL min(LL a, LL b){            return a > b ? b : a;    }    int main(){            LL t, i, j, r, a[101][101], b[101];            LL T;            cin >> T;             while(T --){                    cin >> t;                    for(i = 0; i < t; ++i){                            cin >> b[i] >> b[i + 1];                    }                    for(i = 0; i <= 100; ++i){                            for(j = 0; j <= 100; ++j){                                    a[i][j] = 0;    //全部归零                            }                    }                    for(r = 2; r <= t; ++r){        //求相邻r个矩阵的最优值                            for(i = 1; i + r - 1 <= t; ++i){        //从第i个矩阵开始的r个矩阵                                    LL num = 0xffffffffffffff;     //定义一个较大的值                                    for(j = i; j < i + r - 1; ++j){ //从第i个矩阵开始的r个矩阵拆分为i到j和j+1到i+r-1两个部分                                            num = min(num, a[i][j] + a[j + 1][i + r - 1] + b[i - 1] * b[j] * b[i + r - 1]);                                    }                                    a[i][i + r - 1] = num;                                    //cout << i << ":" << i + r - 1 << " " << a[i][i + r - 1] << " ";                            }                            //cout <<  endl;                    }                    cout << a[1][t] << endl;            }            return 0;    }

动态规划：

具体见：http://www.clanfei.com/2012/05/1159.html

最后的求最优解的方法值得看看！

12.http://acm.cs.ecnu.edu.cn/problem.php?problemid=1053

#include <stdio.h>#include <iostream>using namespace std;int count,n,x;int s[105];void search(int i,int sum){    if(i>=n)    {        if(sum==x)        {            count++;        }    }    else    {        if(sum+s[i]<=x)        {                  search(i+1,sum+s[i]);        }        search(i+1,sum);    }}int main(){    int T;    cin >> T;    while(T --)    {        cin >> n >> x;        for(int i=0;i<n;i++)        {            cin >> s[i];        }        count = 0;        search(0,0);        cout << count << endl;    }}

NP，T^T,没看懂

13.http://acm.cs.ecnu.edu.cn/problem.php?problemid=1054

#include<stdio.h>int i,j,C,n,tw,tc,bestw;int w[52][3],c[52][3];void backtrack(int i){int j;if(i>=n){if(tw <bestw||bestw==-1){bestw=tw; }return;}for(j=0;j<3;j++){if(tc+c[i][j]<C){tc+=c[i][j];tw+=w[i][j];backtrack(i+1);tc-=c[i][j];tw-=w[i][j];}}} int main(){while(scanf("%d%d",&n,&C)!=EOF){tw=tc=0;bestw=-1;for(i=0;i<n;i++)for(j=0;j<3;j++)scanf("%d%d",&w[i][j],&c[i][j]);backtrack(0);printf("%d\n",bestw);}}

也就EOJ能这么做了

14.http://acm.cs.ecnu.edu.cn/problem.php?problemid=1111

#include <stdio.h>#include <stdlib.h>#include <ctype.h>#include <string.h>#define MAX_ARRAY_SIZE 105int data[MAX_ARRAY_SIZE][MAX_ARRAY_SIZE];//存放最初的数据int cost[MAX_ARRAY_SIZE][MAX_ARRAY_SIZE];//存放各个子问题的最优解int mark[MAX_ARRAY_SIZE][MAX_ARRAY_SIZE];//存放输出最优解方案标志int main(int argc,char *argv[]){   int T;    scanf("%d",&T);    while(T --){        //初始化data        int n;        scanf("%d",&n);        int i,j;        for(i=0;i<n;i++){            for(j=0;j<=i;j++){                scanf("%d",&data[i][j]);            }        }        //dp初始化        for(i=0;i<n;i++){            cost[n-1][i]=data[n-1][i];        }        //dp运行        for(i=n-2;i>=0;i--){            for(j=0;j<=i;j++){                if(cost[i+1][j]<cost[i+1][j+1]){                    cost[i][j]=data[i][j]+cost[i+1][j];                    mark[i][j]=0;                }else{                    cost[i][j]=data[i][j]+cost[i+1][j+1];                    mark[i][j]=1;                }            }        }        printf("%d\n",cost[0][0]);    }    return 0;}

这题一点都不想写，直接copy的人家的代码

15：http://acm.cs.ecnu.edu.cn/problem.php?problemid=1114

#include <cstdio>#include <deque>#include <set>#include <string>#include <map>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <queue>using namespace std; typedef long long LL;const int maxn = 40;bool isprime[maxn];bool visited[22];void init(){    memset(isprime,true,sizeof(isprime));    isprime[0] = false;    isprime[1] = false;    for(int i=2;i<maxn;i++){        if(isprime[i]){            for(int j=2;j<=maxn/i;j++){                isprime[i*j] = false;            }        }    }}int n;int a[22];void dfs(int i){    if(i > n){        if(isprime[a[i-1]+a[1]] == true){            for(int j=1;j<=n;j++){                if(j != 1)printf(" ");                printf("%d",a[j]);            }            printf("\n");            return ;        }    }    for(int j=1;j<=n;j++){        if(visited[j] == true)continue;        if(i!=1 && isprime[a[i-1]+j] == false)continue;        a[i] = j;        visited[j] = true;        dfs(i+1);        visited[j] = false;    }}int main(){    init();    while(cin >> n){        memset(visited,false,sizeof(visited));        a[1] = 1;        visited[1] = true;        dfs(2);    }    return 0;}

贴了那么多代码，还是自己写的看的清楚啊，高端霸气上档次！

7-28 ：15：30

先不刷水题了，没意思，还是看看怎么搞点没接触过的。

先从高斯消元开始吧

16：